\begin{proof}
Suppose we are given a graphical game with degree 3 for which we want to find a personalized equilibrium. We
first convert this graph so that it still has degree 3 but also obeys
the following property: 

\begin{property} \label{prop:transitive_edges}
For each node $u$ with at least two outgoing
edges, one to node $v_1$ and the other to node $v_2$, there exists
either an edge from $v_1$ to $v_2$ or an edge from $v_2$ to $v_1$.
\end{property}

Given a graph with maximum degree 3, we make the following
modifications to satisfy Property \ref{prop:transitive_edges}.  Suppose we have a node
$u$ with an edge to $v_1$ and an edge to $v_2$, and suppose $v_1$
already has degree $3$. To fix this, we will create an extra node
$v_1'$ with the same strategies as $v_1$ which will play exactly the
same weights as $v_1$ in an equilibrium. We can do this by setting the
payoff to $v_1'$ to $1$ whenever $v_1'$ agrees with$v_1$, and setting the payoffs to $v_1'$ to $0$ whenever $v_1'$ disagrees with $v_1$.
Since $v_1'$ only
depends on $v_1$, $v_1'$ has out-degree 1. We have added one to the in-degree of
$v_1$. However, now we can replace edge $(u, v_1)$ with edge $(u,
v_1')$, so $v_1$ now has degree $3$, as originally, and $v_1'$ has
degree $2$. Thus we can add an edge between $v_1'$ and $v_2$ without
exceeding the degree requirement on $v_1$.  Repeating the above
transformations with other vertices that violate the desired property
will lead to a degree-3 graphical game that satisfies Property \ref{prop:transitive_edges}.

Now we have a degree $3$ graph with Property \ref{prop:transitive_edges}. Create a
3-coloring (possible for any degree $3$ graph) and create one player per color. Each player takes all of the
strategies for each node in that color. For ease of notation, assume
that each of the original nodes had only 2 strategies. This can be
easily adjusted for more strategies.  Add dummy strategies as
necessary so that each of the 3 players has the same number of
strategies. Also add a fourth player with half the number of
strategies as any other player.

% using extra $ to fix line break problems.
This gives us 4 players. Call the strategies for player 1 $\{a_{10}$,
$a_{11}$, $a_{20}$, $a_{21}$, $\ldots$, $a_{k0}$, $a_{k1}\}$, the strategies for
player 2 $\{b_{10}$, $b_{11}$, $\ldots$, $b_{k0}$, $b_{k1}\}$, the
strategies for player 3 $\{c_{10}$, $c_{11}$, $\ldots$, $c_{k0}$,
$c_{k1}\}$, and the strategies for player 4 $\{d_1$, $d_2$, $\ldots$,
$d_k\}$.

Next we will assign payoffs for each edge. Start by giving each edge the same payoff as in the graphical
game (we can do this because no two nodes influencing the same
strategy are strategies of the same player). Notice that these payoffs
will not depend at all on player 4. Start by assigning payoff $0$ to player $4$ for each strategy combination. 
Let $p_i(w,x,y,z)$ represent the payoff to player $i$ if player 1 plays
$w$, player 2 plays $x$, player 3 plays $y$, player 4 plays $z$. Now we want
to add to these payoffs in order to ensure that each player plays each
strategy pair equally.

Let $M$ be strictly greater than the largest payoff so far. Now,
adjust the payoffs as follows: $p_1(a_{si}, x, y, d_{s}) += M$ (player 1 is playing either strategy from the node numbered $s$, player 4 is playing his $s^{th}$ strategy). 
$p_2(w, b_{si}, y, d_{s}) += M$ (player 2 is playing either strategy from the node numbered $s$, player 4 is playing his $s^{th}$ strategy). 
$p_3(w, x, c_{si}, d_{s}) += M$ (player 3 is playing either strategy from the node numbered $s$, player 4 is playing his $s^{th}$ strategy). 
$p_4(a_{si}, x, y, d_{(s+1)}) += M$ (player 1 is playing either strategy from the node numbered $s$, player 4 is playing strategy $s \bmod k + 1$). 

If $f_i(x)$ = the amount player $i$ plays strategy $x$ then in any
equilibrium we must have (for all $s$)
\begin{eqnarray*}
f_1(a_{s0}) + f_1(a_{s1}) & = & f_4(d_s) \\
f_2(b_{s0}) + f_2(b_{s1}) & = & f_4(d_s) \\
f_3(c_{s0}) + f_3(c_{s1}) & = & f_4(d_s) \\
f_4(d_s)  & = & f_1(a_{(s-1)0}) + f_1(a_{(s-1)1}) \textrm{ for $1 < s \le k$} \\
f_4(d_1)  & = & f_1(a_{k0}) + f_1(a_{k1})\\
\end{eqnarray*}

These equations imply that:
\begin{eqnarray*}
f_1(a_{s0}) + f_1(a_{s1}) & = & f_1(a_{(s-1)0}) + f_1(a_{(s-1)1})  \textrm{ for $s > 0$} \\
f_1(a_{00}) + f_1(a_{01}) & = & f_1(a_{k0}) + f_1(a_{k1})  \textrm{ for $s > 0$} \\
f_2(b_{s0}) + f_2(b_{s1}) & = & f_1(a_{s0}) + f_1(a_{s1}) \\
f_3(c_{s0}) + f_3(c_{s1}) & = & f_1(a_{s0}) + f_1(a_{s1}) \\
\end{eqnarray*}

In other words, given a personalized equilibrium in this game, we can simply
multiply by the number of pairs (nodes) per player to get a personalized 
equilibrium in the graphical game.
\end{proof}